13312001441113 has 4 divisors (see below), whose sum is σ = 13313969805328. Its totient is φ = 13310033076900.

The previous prime is 13312001441111. The next prime is 13312001441153. The reversal of 13312001441113 is 31114410021331.

It is a semiprime because it is the product of two primes, and also a Blum integer, because the two primes are equal to 3 mod 4.

It is a cyclic number.

It is not a de Polignac number, because 13312001441113 - 2¹ = 13312001441111 is a prime.

It is a Duffinian number.

It is not an unprimeable number, because it can be changed into a prime (13312001441111) by changing a digit.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 984171963 + ... + 984185488.

It is an arithmetic number, because the mean of its divisors is an integer number (3328492451332).

Almost surely, 2^{13312001441113} is an apocalyptic number.

It is an amenable number.

13312001441113 is a deficient number, since it is larger than the sum of its proper divisors (1968364215).

13312001441113 is an equidigital number, since it uses as much as digits as its factorization.

13312001441113 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 1968364214.

The product of its (nonzero) digits is 864, while the sum is 25.

Adding to 13312001441113 its reverse (31114410021331), we get a palindrome (44426411462444).

The spelling of 13312001441113 in words is "thirteen trillion, three hundred twelve billion, one million, four hundred forty-one thousand, one hundred thirteen".